volumes of revolution
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math 131 application: volumes of revolution, part ii 6 6.2 Volumes of Revolution: The Disk Method One of the simplest applications of integration (Theorem 6.1)—and the accumula- tion process—is to determine so-called volumes of revolution. In this section we will concentrate on a method known as the disk method. Solids of Revolution If a region in the plane is revolved about a line in the same plane, the resulting object is a solid of revolution, and the line is called the axis of revolution. The following situation is typical of the problems we will encounter. Solids of Revolution from Areas Under Curves. Suppose that y = f ( x) is a contin- uous (non-negative) function on the interval [ a, b]. Rotate the region under the f between x = a and x = b around the the x-axis and determine the volume of the resulting solid of revolution. See Figure 6.10 f (x i ) a x i b .............................................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . y = f (x) • f (x i ) a x i b ..................................................................................... ..................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • Figure 6.10: Left: The region under the continuous curve y = f (x) on the interval [ a, b]. Right: The solid generated by rotating the region about the x-axis. Note: The point (x i , f (x i )) on the curve traces out a circular cross- section of radius r = f (x i ) when rotated. Once we know the cross-sectional areas of the solid, we can use Theorem 6.1 to determine the volume. But as Figure 6.10 shows, when the point ( x i , f ( x i )) on the curve is rotated about the x-axis, it forms a circular cross-section of radius R = f ( x i ). Therefore, the cross-sectional area at x i is A( x i )= p R 2 = p[ f ( x i )] 2 . Since f is continuous, so is p[ f ( x)] 2 and consequently Theorem 6.1 applies. Volume of Solid of Revolution = Z b a A( x) dx = Z b a p[ f ( x)] 2 dx. Of course, we could use this same process if we rotated the region about the y-axis and integrated along the y-axis. We gather these results together and state them as a theorem. THEOREM 6.2 (The Disk Method). If V is the volume of the solid of revolution determined by rotating the continuous function f (x) on the interval [ a, b] about the x-axis, then V = p Z b a [ f (x)] 2 dx. (6.2) If V is the volume of the solid of revolution determined by rotating the continuous function f (y) on the interval [c, d] about the y-axis, then V = p Z d c [ f (y)] 2 dy. (6.3) Another Development of the Disk Method Using Riemann Sums Instead of using Theorem 6.1, we could obtain Theorem 6.2 directly by using the ‘subdivide and conquer’ strategy once again. Since we will use this strategy in later situations, let’s quickly go through the argument here. 1 1 There are often several ways to prove a result in mathematics. I hope one of these two will resonate with you.
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volumes of revolution
Transcript of volumes of revolution
math 131 application: volumes of revolution, part ii 6
6.2 Volumes of Revolution: The Disk Method
One of the simplest applications of integration (Theorem 6.1)—and the accumula-tion process—is to determine so-called volumes of revolution. In this section wewill concentrate on a method known as the disk method.
Solids of Revolution
If a region in the plane is revolved about a line in the same plane, the resultingobject is a solid of revolution, and the line is called the axis of revolution. Thefollowing situation is typical of the problems we will encounter.
Solids of Revolution from Areas Under Curves. Suppose that y = f (x) is a contin-uous (non-negative) function on the interval [a, b]. Rotate the region under the fbetween x = a and x = b around the the x-axis and determine the volume of theresulting solid of revolution. See Figure 6.10
f (xi)
a xi b
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y = f (x)• f (xi)
a xi b
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................................ •Figure 6.10: Left: The region underthe continuous curve y = f (x) onthe interval [a, b]. Right: The solidgenerated by rotating the region aboutthe x-axis. Note: The point (xi , f (xi))on the curve traces out a circular cross-section of radius r = f (xi) whenrotated.
Once we know the cross-sectional areas of the solid, we can use Theorem 6.1to determine the volume. But as Figure 6.10 shows, when the point (xi, f (xi)) onthe curve is rotated about the x-axis, it forms a circular cross-section of radiusR = f (xi). Therefore, the cross-sectional area at xi is
A(xi) = pR2 = p[ f (xi)]2.
Since f is continuous, so is p[ f (x)]2 and consequently Theorem 6.1 applies.
Volume of Solid of Revolution =Z b
aA(x) dx =
Z b
ap[ f (x)]2 dx.
Of course, we could use this same process if we rotated the region about the y-axisand integrated along the y-axis. We gather these results together and state them asa theorem.
THEOREM 6.2 (The Disk Method). If V is the volume of the solid of revolution determined byrotating the continuous function f (x) on the interval [a, b] about the x-axis, then
V = pZ b
a[ f (x)]2 dx. (6.2)
If V is the volume of the solid of revolution determined by rotating the continuous functionf (y) on the interval [c, d] about the y-axis, then
V = pZ d
c[ f (y)]2 dy. (6.3)
Another Development of the Disk Method Using Riemann Sums
Instead of using Theorem 6.1, we could obtain Theorem 6.2 directly by using the‘subdivide and conquer’ strategy once again. Since we will use this strategy inlater situations, let’s quickly go through the argument here.1 1 There are often several ways to prove
a result in mathematics. I hope one ofthese two will resonate with you.
math 131 application: volumes of revolution, part ii 7
As above, we start with a continuous function on [a, b]. This time, though, wecreate a regular partition of [a, b] using n intervals and draw the correspondingapproximating rectangles of equal width Dx. In left half of Figure 6.11 we havedrawn a single representative approximating rectangle on the ith subinterval.
a Dx b
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f (xi)
y = f (x)
a Dx b
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y = f (x)........................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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Figure 6.11: Left: The region underthe continuous curve y = f (x) onthe interval [a, b] and a representativerectangle. Right: The disk (cylinder) ofradius R = f (xi) generated by rotatingthe representative rectangle aboutthe x-axis. The volume or this disk ispR2w = p[ f (xi)]2Dx.
Rotating each representative rectangle creates a representative disk (cylinder) ofradius R = f (xi). (See the right half of Figure 6.11.) The volume of this cylinder isgiven by (6.1)
Volume of a Cylinder = (area of the base)⇥ height.
In this case when the disk is situated on its side, we think of the height as the‘width’ Dx of the disk. Moreover, since the base is a circle, its area is pR2 =
p[ f (xi)]2 so
Volume of a representative disk = DVi = p[ f (xi)]2Dx.
To determine the volume of entire solid of revolution, we take each approximat-ing rectangle, form the corresponding disks (see the middle panel of Figure 6.12)and sum the resulting volumes, it generates a representative disk whose volume is
DV = pR2Dx = p[R(xi)]2Dx.
Figure 6.12: A general solid of revolu-tion and its approximation by a seriesof n disks. (Diagram from Larson &Edwards)
Approximating the volume of the entire solid by n such disks (see the right-hand panel of Figure 6.12) of width Dx and radius f (xi) produces a Riemann sum
Volume of Revolution ⇡n
Âi=1
p[ f (xi)]2Dx = p
n
Âi=1
[ f (xi)]2Dx. (6.4)
As usual, to improve the approximation we let the number of subdivisions n ! •and take a limit. Recall from our earlier work with Riemann sums, this limit exists
math 131 application: volumes of revolution, part ii 8
because [ f (x)]2 is continuous on [a, b] since f (x) is continuous there.
Volume of Revolution = limn!•
pn
Âi=1
[ f (xi)]2Dx = p
Z b
a[ f (x)]2 dx (6.5)
where we have used the fact that the limit of a Riemann sum is a definite integral.This is the same result we obtained in Theorem 6.2. Ee could use this same processif we rotated the region about the y-axis and integrated along the y-axis.
Stop! Notice how we used the ‘subdivide and conquer’ process to approximatethe quantity we wish to determine. That is we have subdivided the volume into‘approximating disks’ whose volume we know how to compute. We have thenrefined this approximation by using finer and finer subdivisions. Taking the limitof this process provides the answer to our question. Identifying that limit with anintegral makes it possible to easily (!) compute the volume in question. OK, timefor some examples.
I’ll admit it is hard to draw figures like Figure 6.12. However, drawing a rep-resentative rectangle for the region in question, as in the left half of Figure 6.11 isusually sufficient to set up the required volume integral.
Examples
Let’s start with a couple of easy ones.
EXAMPLE 6.3. Let y = f (x) = x2 on the interval [0, 1]. Rotate the region between thecurve and the x-axis around the x-axis and find the volume of the resulting solid.
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f (x) = x2
1
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f (x) = x2
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Figure 6.13: Left: A representativerectangle for the curve y = x2. Right:A representative circular slice for thecurve y = x2 rotated about the x-axis.
SOLUTION. Using Theorem 6.2
V = pZ b
a[ f (x)]2 dx = p
Z 1
0[x2]2 dx =
px5
5
����1
0=
p
5.
Wow, that was easy!
EXAMPLE 6.4. Let y = f (x) = x2 on the interval [0, 1]. Rotate the region between thecurve and the y-axis around the y-axis and find the volume of the resulting solid.
0
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f (x) = x2
1
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py
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Figure 6.14: Left: A representativerectangle for the region between thecurve y = x2 (x =
py) and the y-axis.
Right: A representative circular slicefor the curve x =
py rotated about the
x-axis.
SOLUTION. The region is not the same one as in Example 6.3. It lies between they-axis and the curve, not the x-axis. See Figure 6.14.
Since the rotation is about the y-axis, we need to solve for x as a function of y.Since y = x2, then x =
py. Notice that the region lies over the interval [0, 1] on the
y-axis now. Using Theorem 6.2
V = pZ d
c[g(y)]2 dy = p
Z 1
0[p
y]2 dy =py2
2
����1
0=
p
2.
EXAMPLE 6.5. Find the volume of a sphere of radius r which can be obtained byrotating the semi-circle f (x) =
pr2 � x2 about the x-axis.
�r r
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Figure 6.15: Left: A representativerectangle for the curve y =
pr2 � x2.
Right: A representative circular slice forthe sphere that results when rotatingthe semi-circle about the x-axis.
math 131 application: volumes of revolution, part ii 9
SOLUTION. Using Theorem 6.2
V = pZ b
a[ f (x)]2 dx = p
Z r
�r[p
r2 � x2]2 dx
= pZ r
�rr2 � x2 dx
= p
✓r2x � x3
3
◆ ����r
�r
= p
✓r3 � r3
3
◆�
✓�r3 +
r3
3
◆�
=4pr3
3.
Amazing! We have derived the volume formula of a sphere from the volume by disksformula.
YOU TRY IT 6.9. Find the volume of a cone of radius r and height h by rotating the linethrough (0, 0) and (r, h) about the y-axis. (See Figure 6.16.)
r
h
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Figure 6.16: Determine the volume of acone of radius r and height h.
EXAMPLE 6.6 (Two Pieces). Consider the region enclosed by the curves y =p
x, y = 6 �x, and the x-axis. Rotate this region about the x-axis and find the resulting volume.
SOLUTION. It is important to sketch the region to see the relationship between thecurves. Pay particular attention to the bounding curves. Draw representative rectan-gles. This will help you set up the appropriate integrals. It is less important to try todraw a very accurate three-dimensional picture.
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Figure 6.17: Left: The region enclosedby the curves y =
px, y = 6 � x,
and the x-axis and two representativerectangles. Right: The resulting solid ofrevolution about the x-axis is formedby two distinct pieces each requiring itsown integral.
First determine where the curves intersect: Obviously y =p
x meets the x-axis atx = 0 and y = 6 � x meets the x-axis at x = 6. For y = 6 � x and y =
px,
6 � x =p
x ) 36 � 12x + x2 = x ) x2 � 13x + 36 = (x � 4)(x + 9) = 0
) x = 4, (x = �9 is not in the domain).
From Figure 6.17 we see that the solid is made up of two separate pieces (the topcurve changes at x = 4) and each requires its own integral. Using Theorem 6.2
V = pZ 4
0[p
x]2 dx + pZ 6
4[6 � x]2 dx
= pZ 4
0x dx + p
Z 6
4[6 � x]2 dx
=px2
2
����4
0+
�p(6 � x)3
3
����6
4
= (8p � 0) +✓
0 � (�8p)3
◆
=32p
3.
Note: We used a ‘mental adjustment’ to do the second integral (with u = 6 � x anddu = �dx). Overall, the integration is easy once the problem is set up correctly. Besure you have the correct region.
math 131 application: volumes of revolution, part ii 10
EXAMPLE 6.7 (One Piece: Two integrals). Reconsider the same region as in Example 6.6enclosed by the curves y =
px, y = 6 � x, and the x-axis. Now rotate this region
about the y-axis instead and find the resulting volume.
SOLUTION. OK, the region is the same as above. Here is where you have to be verycareful. Since the rotation is about the y-axis, the strips are horizontal this time. No-tice that there are two strips. When this region is rotated about the y-axis, the solidwill have a ‘hollowed out’ center portion (see Figure 6.18). Thus, we must take theouter region formed by the curve y = 6 � x and subtract the inner region formed byy =
px from it.
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Figure 6.18: Left: The region enclosedby the curves y =
px, y = 6 � x,
and the x-axis and two representativerectangles. Right: The resulting solid ofrevolution about the x-axis (a ‘volcano’)is formed by two distinct pieces eachrequiring its own integral.
Since the rotation is about the y-axis, to use the disk method we need to write thecurves in the form x as a function of y. We have
y =p
x ) x = y2 and y = 6 � x ) x = 6 � y.
The outer curve is x = 6 � y and the inner curve is x = y2 from the perspective of they-axis. From the previous problem we already know the intersection points. Howeverwe need their y-coordinates. When x = 4 the corresponding y-coordinate (using eithery =
px or y = 6 � x) is y = 2. The other coordinate is y = 0. So this time using
Theorem 6.2 (note the subtraction of the inner volume)
V = Outer Volume � Inner Volume
= pZ 2
0[6 � y]2 dy � p
Z 2
0[y2]2 dy
=�p(6 � y)3
3
����2
0� py5
5
����2
0
=
✓�64p
3� �216p
3
◆�
✓32p
5� 0
◆
=664p
15.
Whew! That’s a lot of things to process. Getting a clear picture of the region iscrucial. Even if you never draw the three-dimensional version, understanding therepresentative rectangles is critical. In this case the representative rectangles didnot extend from the bottom to the top curve (i.e., the left to the right), but ratherfrom the y-axis to each curve separately to represent the outer solid and the innersolid.
Caution! We wrote the volume of the solid as the outer minus inner volume:
V = pZ 2
0[6 � y]2 dy � p
Z 2
0[y2]2 dy.
Since the interval is the same for both integrals, we could have written this as
V = pZ 2
0[6 � y]2 � [y2]2 dy
where each individual radius is squared separately. If you do combine the integrals,you cannot square the difference of the two radii:
V 6= pZ 2
0[6 � y � y2]2 dy
math 131 application: volumes of revolution, part ii 11
because[6 � y]2 � [y2]2 6= [6 � y � y2]2.
You’ve been warned!Look, there are two situations you need to distinguish when multiple curves
are used for a solid of revolution. The curves might create two pieces which sumtogether to create the entire solid in which case you will need to add integralstogether to find the volume (see Figure 6.17). Or the curves might be situated sothat resulting solid is formed by hollowing out one solid from another in whichcase you will need to subtract one integral from another (see Figure 6.18). This iswhy a sketch of the position of the curves relative to the axis of rotation is critical.Let’s do a couple more.
EXAMPLE 6.8. Consider the region enclosed by the curves y =p
x, y = x � 2, and thex-axis. Rotate this region about the x-axis and find the resulting volume.
SOLUTION. Is this the sum of two integrals or is it difference of two integrals?First determine where the curves intersect: Obviously y =
px meets the x-axis at
x = 0 and y = x � 2 meets the x-axis at x = 2. Now y = x � 2 and y =p
x intersectwhen
x � 2 =p
x ) x2 � 4x + 4 = x ) x2 � 3x + 4 = (x � 4)(x + 1) = 0
) x = 4, (not x = �1). 0 2 40
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Outer Inner
Figure 6.19: Left: The region enclosedby the curves y =
px, y = 2 � x, and
the x-axis. When rotated about thex-axis, one region must be subtractedfrom the other. Instead of using repre-sentative rectangles, we simply indicatethe appropriate radii with arrows.
From the sketch in Figure 6.19 if revolved about the x-axis will result in a solid thathas been partially hollowed out (a cone has been removed). This requires a differenceof integrals. Using Theorem 6.2
V = Outer Volume � Inner Volume
= pZ 4
0[p
x]2 dx � pZ 4
2[x � 2]2 dx
= pZ 4
0x dx � p
Z 4
2[x � 2]2 dx
=px2
2
����4
0� p(x � 2)3
3
����4
2
= (8p � 0) +✓
8p
3� 0
◆
=16p
3.
EXAMPLE 6.9. Again consider the region enclosed by the curves y =p
x, y = x � 2,and the x-axis. This time rotate the region about the y-axis and find the resultingvolume.
SOLUTION. Is this the sum of two integrals or is it difference of two integrals? Sincethe rotation is about the y-axis, the radii of the respective regions are horizontal, seeFigure 6.20. This is again a difference of two integrals.
Translating the curves into functions of y we have x = y2, x = y + 2, and y = 0 (thex-axis). The curves intersect the x-axis at y = 0. We’ve seen that the line and squareroot function meet when x = 4 since x = y + 2 there, then the y-coordinate of theintersection is y = 2. 0 2 4
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Inner
Figure 6.20: The region enclosed by thecurves y =
px, y = 2 � x, and the
x-axis and representative radii. Whenrotated about the y-axis, one regionmust be subtracted from the other.
math 131 application: volumes of revolution, part ii 12
Using Theorem 6.2
V = Outer Volume � Inner Volume = pZ 2
0[y + 2]2 dy � p
Z 2
0[y2]2 dy
=p(y + 2)3
3
����2
0� py5
5
����2
0
=
✓64p
3� 8p
3
◆�
✓32p
5� 0
◆
=184p
15.
EXAMPLE 6.10. Consider the region enclosed by the curves y = ex, x = 1 and they-axis. Rotate the region about the y-axis and find the resulting volume.
SOLUTION. Is this the sum of two integrals or is it difference of two integrals? Sincethe rotation is about the y-axis, the radii of the respective regions are horizontal, seeFigure 6.21. This is again a difference of two integrals.
Translating the curves into functions of y we see y = ex becomes x = ln y. Noticethat x = ln y meets the y-axis at 1 and it meets the line x = 1 at y = e.
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Outer
Inner
y = ex orx = ln y
Figure 6.21: The region enclosed by thecurves y = ex , x = 1 and the y-axis.Rotate the region about the y-axis andfind the resulting volume.
The solid formed by the region between the y-axis and the curve x = ln y mustbe subtracted from the ordinary cylinder formed by rotating the line x = 1 about they-axis. This cylinder has radius 1 and height e so its volume is v = p(1)2e = pe. (Wecould also find this by integrating, but why bother.) So Using Theorem 6.2
V = Outer Volume � Inner Volume = pe � pZ e
1[ln y]2 dy =???
Since we don’t know an antiderivative for (ln x)2 we are stuck. (By the way, noticethat the limits of integration for the integral are from 1 to e, not 0 to e.) So we willhave to invent another method of finding such volumes if we are to solve this prob-lem. This shows the importance of having a wide-variety of methods for solving asingle problem.
YOU TRY IT 6.10. Let R be the entire region enclosed by y = x2 and y = 2 � x2 in the upperhalf-plane. Sketch the region. Rotate R about the x-axis and find the resulting volume.(Answer: 16
3 p.)
YOU TRY IT 6.11. Let R be the region enclosed by y = 2x and y = x2.
(a) Rotation about the x-axis. Find the volume of the hollowed-out solid generated byrevolving R about the x-axis. (Answer: 64p/15)
(b) Rotation about the y-axis. Find the volume of the solid generated by revolving Rabout the y-axis by using the disk method and integrating along the y-axis. (Answer:8p/3)
math 131 application: volumes of revolution, part ii 13
6.3 Rotation About Other Axes
It is relatively easy to adapt the disk method to finding volumes of solids of rev-olution using other horizontal or vertical axes. The key steps are to determine theradii of the slices and express them in terms of the correct variable. A couple ofexamples should give you the idea.
EXAMPLE 6.11. Consider the region enclosed by the curves y = x2 � 2x and y = 3.Rotate this region about the line y = 3 and find the resulting volume.
SOLUTION. The two curves meet when
x2 � 2x = 3 ) x2 � 2x � 3 = (x � 3)(x + 1) = 0 ) x = 3, �1.
The parabola and the line are easy to sketch; see Figure 6.22 on the left.
�1 3
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y = 3
y = x2 � 2x
�1 3
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............................................................................................................................................................ Figure 6.22: Left: The region enclosed
by the curves y = x2 � 2x and y = 3.Since the axis of revolution is y = 3, arepresentative radius extends from theline y = 3 to the curve y = x2 � 2x.Right: The resulting solid of revolutionabout the line y = 3.
A representative radius extends from the line y = 3 to the curve y = x2 � 2x. Thelength of a radius is the difference between these two values, that is, the radius of acircular cross-section perpendicular to the line y = 3 is
r = 3 � (x2 � 2x) = 3 + 2x � x2.
Since a cross-section is a circle, its area is
A = A(x) = pr2 = p(3 + 2x � x2)2.
Since we know the cross-sectional area, we can use Theorem 6.1 to find the volume
V =Z b
aA(x) dx = p
Z 3
�1(3 + 2x � x2)2 dx
= pZ 3
�19 + 12x � 2x2 � 4x3 + x4)2 dx
= p
✓9x + 6x2 � 2x3
3� x4 +
x5
5
◆ ����3
�1
= p
✓27 + 54 � 18 � 81 +
2435
◆�
✓�9 + 6 +
23� 1 � 1
5
◆�
=512p
15.
EXAMPLE 6.12. Consider the region enclosed by the curves x = y2 and x = 2 � y2.Rotate this region about the line x = 3 and find the resulting volume.
SOLUTION. The two curves meet when
y2 = 2 � y2 ) 2y2 = 2 ) y = ±1.
The region is easy to sketch; see Figure 6.23 on the left.An ‘outer’ representative radius extends from the line x = 3 to the curve x = y2
and an ‘inner’ representative radius extends from the line x = 3 to x = 2 � y2. Thelength of a representative radius is the difference between the corresponding pairs of
math 131 application: volumes of revolution, part ii 14
�1
1.......................................................................................................................................................................................................................................................
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x = 3
x = y2
x = 2 � y2
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�1
1 ..........................................................................................................................................................................................
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x = 3
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..
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............. ............. ............. ............. ............. .........................................................................
.......
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...... Figure 6.23: Left: The region enclosedby the curves x = y2 and x = 2 � y2
and two representative radii emanatingfrom the axis of rotation x = 3. Right:The resulting ‘hollowed out’ solid ofrevolution about the line x = 3 lookslike a ‘doughnut.’
values. The outer radius is R = 3 � y2 and the inner radius is 3 � (2 � y2) = 1 + y2.The integration will take place along the y-axis on the interval [�1, 1] because thedisks are horizontal when points are rotated about the line x = 3. Since we know thecross-sections are circles, we can use Theorem 6.1 to find the volume
V = Outer Volume � Inner Volume = pZ 1
�1(3 � y2)2 dy � p
Z 1
�1(1 + y2)2 dy
= pZ 1
�19 � 6y2 + y4 dy � p
Z 1
�11 + 2y2 + y4 dy
= pZ 1
�18 � 8y2 dy
= p
✓8y � 8y3
3
◆ ����1
�1
= p
✓8 � 8
3
◆�
✓�8 +
83
◆�
=32p
3.
Because the interval of integration was the same for both the inner and outer volumes,we are able to combine the two integrals which greatly simplified the integration andevaluation. If you understand this problem, you are in good shape.
YOU TRY IT 6.12. Let R be the region enclosed by y = x3, x = 2, and the x-axis. Draw theregion.
(a) Rotate R about the x-axis and find the resulting volume. (Answer: 128p/7)
(b) Rotate R about the line y = 9 and find the resulting volume. (Answer: 376p/7)
(c) Rotate R about the line x = 2 and find the resulting volume. (Answer: 16p/5)
YOU TRY IT 6.13. Draw the region R in the first quadrant enclosed by y = x2, y = 2 � x andthe x-axis.
(a) Rotate R about the x-axis and find the resulting volume. [Is it the sum of two integralsor outside minus inside?]
(b) Rotate R about the y-axis and find the resulting volume. [Is it the sum of two integralsor outside minus inside?]
YOU TRY IT 6.14. Let S be the region in the first quadrant enclosed by y = x2, y = 2 � x andthe y-axis.
(a) Rotate S about the x-axis. [Is it the sum of two integrals or outside minus inside?]
(b) Rotate S about the y-axis. [Is it the sum of two integrals or outside minus inside?]
YOU TRY IT 6.15. Let R be the region in the upper half-plane bounded by y =p
x + 2, thex-axis and the line y = x. Find the volume resulting when R is rotated around the x-axis.Remember: Outside minus inside. How many integrals do you need?
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.........
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Figure 6.24: The region for you try it
6.15 .
YOU TRY IT 6.16. A small canal bouy is formed by taking the region in the first quadrantbounded by the y-axis, the parabola y = 2x2, and the line y = 5 � 3x and rotating it aboutthe y-axis. (Units are feet.) Find the volume of this bouy. (Answer: 2p cubic feet.)
math 131 application: volumes of revolution, part ii 15
YOU TRY IT 6.17. Just set up the integrals for each of the following volume problems. Sim-plify the integrands where possible. Use figures below.
(a) R is the region enclosed by y = x2, y = x + 2, and the y-axis in the first quadrant.Rotate R about the x-axis.
(b) Same R, but rotate around the y-axis.
(c) Same R, but rotate around the line y = 4.
(d) S is the region enclosed by y = �x2 + 2x + 3, y = 3x � 3, the y-axis, and the x-axis inthe first quadrant. Rotate S about the x-axis.
(e) T is the region enclosed by y = �x2 + 2x + 3, y = 3x � 3, and the x-axis in the firstquadrant. Rotate T about the x-axis.
(f ) R is the region enclosed by y = 12 x2 + 2 and y = x2. Rotate R about the x-axis.
(g) Same R, but rotate around the line y = 4.
(h) S is the region enclosed by y = 12 x2 + 2 and y = x2 in the first quadrant. Rotate S
about the y-axis.
(i) T is the region enclosed by y =p
x � 2, y = 4 � x, the y-axis and the x-axis. Rotate Tabout the y-axis.
(j) V is the region enclosed by y =p
x � 2, y = 4 � x, and the x-axis. Rotate V about they-axis.
(k) Same V, but rotate around the x-axis.
(l) Hard: Same V, but rotate around the line y = 2.
0
2
4
6
2 4 6
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...............................................................................................................................................................................................................................................................................................................................................a–c
R
0
2
4
6
2 4 6
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d–e
S
T
2
4
6
�3 �1 1 3
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RR
f–g
2
4
6
0 2
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S
h
0
2
4
6
2 4 6
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................................
.........
i–l
T
V
Figure 6.25: The regions for you try
it 6.17 .
SOLUTION. Here are the solutions before the integands have been simplified. Re-member, you should simplify each integrand before actually integrating.
(a) pZ 2
0(x + 2)2 dx � p
Z 2
0(x2)2 dx (b) p
Z 4
0(p
y)2 dy � pZ 4
2(y � 2)2 dy
(c) pZ 2
0(4 � x2)2 dx � p
Z 2
0(4 � (x + 2))2 dx (d) p
Z 2
0(�x2 + 2x + 3)2 dx � p
Z 2
1(3x � 3)2 dx
(e) pZ 2
1(3x � 3)2 dx + p
Z 3
2(�x2 + 2x + 3)2 dx (f ) Homework
math 131 application: volumes of revolution, part ii 16
(g) Homework (h) pZ 4
0(p
y)2 dy � pZ 4
2
⇣p2y � 4
⌘2dy
(i) pZ 1
0(y2 + 2)2 dy + p
Z 4
1(4 � y)2 dy
(j) pZ 1
0(4 � y)2 dy � p
Z 1
0(y2 + 2)2 dy (k) p
Z 3
2(p
x � 2)2 dx + pZ 4
3(4 � x)2 dx
(l) pZ 3
222 dx � p
Z 3
2(2 �
px � 2)2 dx +
pZ 4
322 dx � p
Z 4
322(2 � (4 � x))2 dx
�
YOU TRY IT 6.18 (Rotation about the x-axis). Let R be the region in the first quadrant enclosedby y =
px � 1, y = x � 7 and the x-axis. Sketch the region. Rotate R about the x-axis and
find the resulting volume. (Answer: 63/2p.)
YOU TRY IT 6.19. Let R be the entire region enclosed by y = x2 and y = 2 � x2 in the upperhalf-plane. Sketch the region. Rotate R about the x-axis and find the resulting volume.(Answer: 16
3 p.)
YOU TRY IT 6.20 (Rotation about the y-axis). Let R be the region enclosed by the x-axis, y =px, and y = 2 � x. Rotate R about the y-axis and find the volume. (Answer: 32p/15.)
YOU TRY IT 6.21 (Rotation about a line parallel to the x-axis). Find the volume of the hollowed-out solid generated by revolving R about the line y = 4. (Answer: 32p/5)
YOU TRY IT 6.22 (Rotation about a line parallel to the x-axis). Let R be the region enclosed byy = 1 � x2 and the x-axis. Rotate R about the line y = 2 and find the resulting volume.(Answer: 64p/15.)
YOU TRY IT 6.23. Extra Fun.
(a) Let R be the region enclosed by y = arctan x, y = p/4, and the y-axis. Find thearea of R. Hint: Only integration along one axis is possible at this point. (Answer:ln
p2 = 0.5 ln 2)
(b) Rotate R about the y-axis and find the resulting volume. Use a trig identity for tan2 q.(Answer: p � p2/4)
YOU TRY IT 6.24. Here are a few more:
(a) Let y = 1x on [1, a], where a > 1. Let R be the region under the curve over this interval.
Rotate R about the x-axis. What value of a gives a volume of p/2? (Answer: a = 2)
(b) What happens to the volume if a ! •? Does the volume get infinitely large? Uselimits to answer the question.
(c) Instead rotate R about the y-axis. What value of a gives a volume of p/2? This isparticularly easy to do by shells! (Answer: a = 5/4)
(d) Instead rotate R about the line y = �1. If we let a = e what is the resulting volume?(Answer: 3p � p
e )
YOU TRY IT 6.25. Let R be the region enclosed by y =p
x, y = 2, and the y-axis.
(a) Rotate R about the x-axis and find the resulting volume. (Answer: 8p)
(b) Rotate R about the line y = 2 and find the resulting volume. (Answer: 8p/3)
(c) Rotate R about the line y = 4 and find the resulting volume. (Answer: 40p/3)
